A bullet is fired from at a shooting range. The bullet hits the ground after 0.3
ID: 2123989 • Letter: A
Question
- A bullet is fired from at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s?
- A not so brilliant physics students wants to jump from their 3rd floor apartment window to the swimming pool below. The problem is the base of the apartment is 8.00 meters from the pool%u2019s edge. If the window is 20.0 meters high, how fast does the student have to be running horizontally to make sure they make it to the pool%u2019s edge?
- If a 1500 kg car stopped from an in 5.6 seconds with an applied force of 5000 N, how fast was it initially traveling?
- If the acceleration due to gravity on the Moon is 1/6 that what is on the Earth, what would a 100 kg man weight on the Moon? If a person tried to simulate this gravity in an elevator, how fast would it have to accelerate and in which direction?
- A 7.93 kg box is pulled along a horizontal surface by a force F_P of 84.0 N applied at a 47.0o angle. If the coefficient of kinetic friction is 0.35, what is the acceleration of the box?
- If a car is traveling at 50 m/s and then stops over 300 meters (while sliding), what is the coefficient of kinetic friction between the tires of the car and the road?
Explanation / Answer
1)
2) S = ut + 1/2 at^2
since a=0 in horizontal direction
S = ut
= 1100*0.32
S= 352 m
3) The student has to cover 8m horizontal distance before he hits the ground.
Therefore first we have to calculate the time taken to reach ground from 20 m height with only a= g(gravity) acting in the downward direction
In vertical direction,
S= ut+1/2 a t^2
u=0 since initially vertical velocity is 0
S=20 a =g = 9.8 m/s (if g taken as 10ms^-2 results will vary)
t^2 = 2*S /a = 2* 20/ 9.8 =
t = 2.019 s
In horizontal direction,
a=0
S= ut
u= S/t = 8/2.019
u= 3.96 m/s
4)Given F= 5000 N m = 1500 kg t = 5.6 s
u = ? v = 0 , since car stopped finally
F = ma
a = F/m = 10/3 ms^-2
v= u + at
u = at = 3.33 * 5.6 =
u = 18.67 ms^-1
5) Weight = mg
on moon a = g/6
therefore Weight of m=100 kg on Moon is,
W = 100 * 9.8 / 6
W = 163.33 N on moon
To simulate the same situation in a lift
N = W (N is the normal reaction by the lift floor and W is weight on moon )
N + Fp = mg ( Fp is pseudo force , if lift is moving down and mg is force of gravity by earth0
mg/6 + ma = mg
a = g - g/6 = 5g/6 ( since acceleration has turned out to be positive our assumption is right )
=5*9.8 /6
a= 8.167 ms^-2
Hence the lift should move downwards with an acceleration 8.167 ms^-2 , i.e., 5g/6
6) Given,
m = 7.93
F_P = 84.0 N angle O= 47.0o
u = 0.35
a =?
In horzontal direction force is
Balancing forces,
F-p cos 47 - u N = ma ------- (1)( u N is the kinectic friction force u = coeff of kinetic friction , N is Normal reaction )
In vertical direction ,
N + F-p sin 47 = mg
N = mg - F-p sin 47
= 7.93 * 9.8 - 84 *0.73
= 77.714- 61.32
= 16.394 N
Using this in equation (1)
F-p cos 47 - u * (16.394) = ma
84 * 0.68 - 0.35* 16.394 = 7.93 * a
7.93 * a = 57.12 - 5.74 = 51.38
a = 51.38/7.93
a = 6.48 m/s^2
7) Given,
S = 300 m u = 50 m/s v = 0
a =?
v^2 - u^2 = 2aS
a = u^2 / (2S) = - 2500/600 = 4.17 m/s^2 (decelration)
u = ?
F = u N
In vertical direction , N = mg
In horizontal direction,
u N = m a
u * m g = m a
u = a/g = 4.17 / 9.8
u = 0.425 coefficient of kinetic friction
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